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[Helgoland]

Well, what can I say - we're a country of engineers. You try explaining to one of those guys that 99.99% precision is not the same as 100%.

[TheBiggerFish]

Blecht.

Also, Helgo, what part of the || postulate makes the middle line connecting c and d parallel again?  I don't recall anything that would necessarily make that true, though I welcome being proved wrong here.  e's not necessarily collinear with f, at least.

(wish they had used points instead of arbitrary line letters...)

[Helgoland]

e's not necessarily collinear with f

Assuming the person who came up with his shit had even a tiny bit of common sense

If it isn't, you can basically just go hang yourself. There's no way to solve it in that case.

[TheBiggerFish]

Well, no, I THINK you can do something with similar triangles.  Thhhhhhhhhhhnope.   Maybe?  Thinking.  Will return once I have slept.

E:Or not.
How did you find b?

[MagmaMcFry]

Hahaha, the problem is in fact broken.

Consider that f/c = (e+f)/d, so f = 12. Now everybody take a look at the triangle acf, it has sides of length 4, 8 and 12. Tell me, how can a triangle with those side lengths have α = 45°?

[TheBiggerFish]

Exactly!  This figure does not make sense!

[hops]

Again, I have to do this sort of things to get to Germany. I'm guessing I have to cope with it and try not to go insane?

[ZetaX]

One should also wonder why they even give the angle alpha at all. Solvable without it, and self-contradictory with it.

But what exactly does it mean that you have to do this crap to get to Germany? If this is an actual entrance test, I expect them not giving it out as homework, so where does it come from?

[TheBiggerFish]

I just have to question the sanity of the writers of that thing...

[frostshotgg]

Food for thought: A triangle with sides 4, 8, and 12 is not a triangle at all. The sum of any two sides must be greater than the third. To put it another way, since 4+8=12, the 12 side is as long as the other two put together, which means that the only way for all three to touch is if they overlap.

[TheBiggerFish]

@Ipsil:So the answer is, "Your problem is broken."?

In Other News:

Haaaaalp.

I cannot into big-O complexity of radix sort.  I'm not even sure it HAS one.

Feel free to redirect this to the programming help thread.

[frostshotgg]

That's without a doubt a programming question.

[TheBiggerFish]

It's got SOMETHING to do with math...I just don't know WHAT.  Eeeeurgh.  Right, I'll ask over there.

[Spehss _]

Physics question.

Spoiler (click to show/hide)

A crate of mass M=80 kg is pushed at a constant speed up a frictionless ramp at an incline of theta=30^o by a horizontal force F. What are the magnitudes of A) the force F and B) the force on the crate from the ramp?

So I've worked out that the force of the box's weight is W=80*9.8=784N. Since the box is sloped at an angle of 30^o and I'm treating the horizontal axis to be parallel to the slope of the ramp, weight is a vector tilted at an angle of 30^o.

So the x and y components of this vector W is
Wx=784sin30=392N
Wy=784cos30=678.96N or ~679N

Force F is perpendicular to W, so it's also at an angle of 30^o from the horizontal, so it's components is
Fx=Fcos30
Fy=Fsin30

And since both Wy and Fy are pushing in the same direction, down the y axis, by Newton's law it means the ramp will be pushing up in an equal and opposite force,
so Fn=Wy+Fy

So there's 5 forces acting on the box and contributing to the net force. Wy and Fy are pushing in the same direction, Fn=-1(Wy+Fy) and cancelling out Wy and Fy, and Fx and Wx are in opposite directions.
Fnet=Fnetx+Fnety

Fnetx=Fx-Wx
Fnety=(Wy+Fy)-Fn

Fnet=(Fx-Wx)+((Wy+Fy)-Fn)

Since it's being pushed at a constant speed, I would think that the net forces acting on the box must be equal to 0. So Fnet=0.
Fnet=0=(Fx-Wx)+((Wy+Fy)-Fn)

0=Fx-Wx
Wx=Fx
Fx=Wx=392N

0=(Wy+Fy)-Fn
0=(Wy+Fy)-(Wy+Fy)
Wy+Fy=Wy+Fy
679+Fy=679+Fy
679-679=Fy-Fy
0=0

...So I can't solve for Fy that way apparently. The way I solved it the first time I figured that if Fx=Wx then Fy must equal Wy and therefore the force of F=W=784N. But according to my physics teacher that's a wrong answer.

And if F=784 and Fy=Wy then Fn=2Wy=679*2=1358N. And that's wrong too.

I don't know what I'm doing wrong. Fnet must be 0 for the speed to be constant, right? And for Fnet to be 0 all the forces acting on the box must equal each other in magnitude but have opposite directions.

[hops]

I'm pretty sure the force of gravity is always downwards, so W should be vertical, not tilted.

Or I might be missing something.

Edit: Nevermind I misread it.